phys2210 sp12 07 FourierandPDEs · Dirac delta function, PDEs, and Fourier transforms If we have...

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Fourier series

2- 2

Fourier Series PDEs Acoustics & Music Optics & diffraction Geophysics Signal processing Statistics Cryptography ...

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How many of the following are even functions? I: x II: sin(x) III: sin2(x) IV: cos2(x)

A)  None B)  Exactly one of them C)  Two of them D)  Three of them E)  All four of them!

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How many of the following are even functions? I: 3x2-2x4 II: -cos(x) III: tan(x) IV: e2x

A)  None B)  Exactly one of them C)  Two of them D)  Three of them E)  All four of them!

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A) Odd B) Even C) Neither

How would you classify this function?

t

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What can you predict about the a’s and b’s for this f(t)?

A)  All terms are non-zero B) The a’s are all zero C) The b’s are all zero D) a’s are all 0, except a0 E) More than one of the above (or none, or ???)

�

f (t) = an cosnω t + bn sinnω tn=0

∞

∑

f(t)

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What can you say about the a’s and b’s for this f(t)?

A)  All terms are non-zero B) The a’s are all zero C) The b’s are all zero D) a’s are all 0, except a0 E) More than one of the above, or, not enough info...

�


∞

∑

t

When you finish P. 3 of the Tutorial, click in:

f(t)

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What can you say about the a’s and b’s for this f(t)?

A)  All terms are non-zero B) The a’s are all zero C) The b’s are all zero D) a’s are all 0, except a0 E) More than one of the above!

�


∞

∑

t

f(t)

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�

f (t) = bn sin(nω t)n=1

∞

∑

Given an odd (periodic) function f(t),

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I claim (proof coming!) it’s easy enough to compute all these bn’s:

bn =2T

f (t)sin(nωt)0

T

∫�


∞

∑

Given an odd (periodic) function f(t),

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f (t) = an cos(nω t)n=0

∞

∑ + bn sin(nω t)n=1

∞

∑If f(t) is neither even nor odd, it’s still easy:

an =2T

f (t)cos(nωt)dt0

T

∫ (if n ≠ 0)

a0 =1T

f (t)dt0

T

∫

bn =2T

f (t)sin(nωt)dt0

T

∫

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For the curve below (which I assume repeats over and over), what is ω?

�


∞

∑

-3 -2 -1 1 2 3

-2

-1

1

2

A) 1 B) 2 C) π D) 2π E) Something else!

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Let’s zoom in. Can you guess anything more about the Fourier series?

�


∞

∑

- 1.0 - 0.5 0.5 1.0

- 2- 1

1

2

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Does this help? (The blue dashed curve is 2Cos πt.)

�


∞

∑

-1.0 -0.5 0.5 1.0

-2

-1

1

2

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�


∞

∑

where bn =2T

f (t)sin(nωt)0

T

∫But why? Where does this formula for bn come from? It’s “Fourier’s trick”!

RECAP: Any odd periodic f(t) can be written as:

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Fourier’s trick: Thinking of functions as a bit like vectors…

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�

c ⋅ d = ci

i=1

3

∑ di

�

v = vi ˆ e ii=1

3

∑Vectors, in terms of a set of basis vectors:

Inner product, or “dot product”:

�

vi =To find one numerical component of v:

�

v ⋅ ˆ e i

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�

v = vi ˆ e ii=1

3

∑

�


∞

∑

Can you see any parallels?

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Inner product, or “dot product” of vectors:

If you had to make an intuitive stab at what might be the analogous inner product of functions, c(t) and d(t), what might you try? (Think about the large n limit?)

�

c ⋅ d = ci

i=1

n

∑ di

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Inner product, or “dot product” of vectors:

If you had to make an intuitive stab at what might be the analogous inner product of functions, c(t) and d(t), what might you try? (Think about the large n limit?) How about:

�

c(t)d(t)dt∫ ??

�

c ⋅ d = ci

i=1

n

∑ di

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What can you say about

�

sin(ωt)sin(2ωt)dt0

T

∫A)  0 B) positive C) negative D) depends E) I would really need to compute it...

2 4 6 8 10 12

- 1.0

- 0.5

0.5

1.0

2 4 6 8 10 12

- 1.0

- 0.5

0.5

1.0

�

sin(ωt)

�

sin(2ωt)

2-

�

sin(mw t)

2 4 6 8 10 12

- 1.0

- 0.5

0.5

1.0

23

If m>1, what can you guess about

�

sin(ωt)sin(mωt)dt0

T

∫

A)  always 0 B) sometimes 0 C)???

2 4 6 8 10 12

- 1.0

- 0.5

0.5

1.0

�

sin(ωt)

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�

sin(nωt)sin(mωt)dt = 00

T

∫ if n ≠ m

Summary (not proven by previous questions, but easy enough to just do the integral and show this!)

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�

sin(nωt)sin(mωt)dt0

T

∫ = 0 (if n ≠ m)�

ˆ e i ⋅ ˆ e j = 0 (if i ≠ j)

Orthogonality of basis vectors:

What does ...

suggest to you, then?

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�

2T

sin(nωt)sin(mωt)dt0

T

∫ =0 (if n ≠ m)1 (if n = m)⎧ ⎨ ⎩

�

ˆ e i ⋅ ˆ e j =0 (if i ≠ j)1 (if i = j)⎧ ⎨ ⎩

Orthonormality of basis vectors:

�

≡ δ i, j

�

≡ δn,m

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�


∞

∑Functions, in terms of basis functions

To find one numerical component:

�

bn =

�

vi =

�

v = vi ˆ e ii=1

n


To find one numerical component:

�

v ⋅ ˆ e i

�

2T

f (t)sin(nωt)dt0

T

∫ (??)

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�

v = vi ˆ e ii=1

n


To find one numerical component: Fourier’s trick

�

ˆ e j ⋅ v = ˆ e j ⋅ vi ˆ e i

i=1

n

∑

�

= vi ˆ e j ⋅ ˆ e ii=1

n

∑

�

= viδ i, ji=1

n

∑

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�

viδ i, ji=1

n

∑ = ?

�

A) vii=1

n

∑B) viC) v j

D) vnE) Other/none of these?

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e j ⋅v = e j ⋅ vi ei

i=1

n

∑ (Fourier's Trick!)

= vi e j ⋅ eii=1

n

∑

�

= viδ i, ji=1

n

∑

�

= v j D’oh!

v = vi eii=1

n

∑

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�


∞

∑

2T

f (t)sin(mωt)0

T

∫ dt = 2T

bn sin(nω t)sin(mω t)dtn=1

∞

∑0

T

∫

= 2T

bn sin(nω t)sin(mω t)dt0

T

∫n=1

∞

∑

To find one component: Fourier’s trick again “Dot” both sides with a “basis vector” of your choice:

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2T

f (t)sin(mωt)0

T

∫ dt = 2T

bn sin(nω t)sin(mω t)dt0

T

∫n=1

∞

∑

�

= bnδn,mn=1

∞

∑

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�

bn δn,mn=1

∞

∑ = ?

�

A) bnn=1

∞

∑B) bnC) bmD) Other/none of these?

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2T

f (t)sin(mωt)0

T

∫ dt = 2T

bn sin(nω t)sin(mω t)0

T

∫n=1

∞

∑ dt

�

= bm

�

= bnδn,mn=1

∞

∑

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�


∞

∑This Fourier sum is used for functions f(t) with period T=2π/ω, which means f(t+T)=f(t) (for all times t) What can you say about the periodicity of the function sin(nωt) ... A)  It has period T = 2π/(nω), but NOT T = 2π/ω B)  It has period T = 2π/ω, but NOT T = 2π/(nω) C)  It has period T = 2π/ω, and ALSO T = 2π/(nω) D)  It has NEITHER period (not T = 2π/ω, not T = 2π/nω)

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We have 3 weeks of class (20% of the term!) left. Still to come: Dirac delta function, PDEs, and Fourier transforms

If we have time after that, how would you like to wrap up the term? More useful math-y stuff like e.g. A)  Special functions (“Bessel Functions” and/or “Legendre Polynomials”)

or more physics-y stuff like... B) Non-inertial reference frames (fictitious forces, like centrifugal and Coriolis) C) I have something else in mind (that I think the whole class would benefit from!) D) I think Prof Rey and Pollock should choose!

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τ

1/τ

Given this little “impulse” f(t) (height 1/τ, duration τ),

In the limit τ è0, what is

�

f (t)dt ?−∞

∞

∫

A)  0 B) 1 C) ∞ D) Finite but not necessarily 1 E) ??

Challenge: Sketch f(t) in this limit.

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The oscillator response to a step function is What is the behavior at large t? x goes to...

�

x(t) =aω02 1− e

−β (t− t0 ) cos(ω1(t − t0)) −βω1

e−β (t− t0 ) sin(ω1(t − t0))⎡

⎣ ⎢

⎤

⎦ ⎥

�

A) 0 B) ∞C) cos(ω1t)D) a/ω0

2 E) Something different!

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What is the value of

�

x 2δ(x − 2)dx−∞

∞

∫

�

A) 0 B) 2C) 4D) ∞E) Something different!

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�

x 2δ(x)dx−∞

∞

∫

�

A) 0 B) 1C) 2D) 4E) 5

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�

(x 2 +1)δ(x)dx−∞

2

∫

�

A) 0 B) 1C) 2D) 4E) 5

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�

x 2δ(x + 2)dx0

∞

∫

�


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�

x 2δ(x + 2)dx−∞

∞

∫

�


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�

x 2δ(2 − x)dx−∞

∞

∫

�

A) 2 B) - 2C) 4D) − 4E) Something different!

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f (t)δ (t − t0 )dt−∞

∞

∫ = f (t0 )

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Recall that What are the UNITS of (where t is seconds)

f (t)δ (t − t0 )dt−∞

∞

∫ =

�

A) sec B) sec-1

C) unitless D) depends on the units of f(t) E) Something different!

δ (t − t0 )

f (t0 )

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τ

1/τ

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PDEs

Partial Differential Equations

2- 49

What is the general solution to Y’’(y)-k2Y(y)=0 (where k is some real nonzero constant)

A)  Y(y)=A eky+Be-ky

B)  Y(y)=Ae-kycos(ky-δ) C)  Y(y)=Acos(ky) D)  Y(y)=Acos(ky)+Bsin(ky) E)  None of these or MORE than one!

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What is the general solution to X’’(x)+k2X(x)=0

A)  X(x)=A ekx+Be-kx

B)  X(x)=Ae-kxcos(kx-δ) C)  X(x)=Acos(kx) D)  X(x)=Acos(kx)+Bsin(kx) E)  None of these or MORE than one!

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TH TC

I’m interested in deriving Where does this come from? And what is α?

∇2T = 1α 2

∂T∂t

Let’s start by thinking about H(x,t), heat flow at x: H(x,t) = “Joules/sec (of thermal energy) passing to the right through position x”

What does H(x,t) depend on?

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TH TC

H(x,t) = Joules/sec (of thermal energy) passing to the right

What does H(x,t) depend on? Probably boundary temperatures! But, how?

A)  H ~ (TH+TC)/2 B)  H ~ TH - TC (=ΔT) C)  Both but not in such a simple way! D)  Neither/???

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TH TC


What does H(x,t) depend on? Perhaps Δx? But, how?

A)  H ~ ΔT Δx B)  H ~ ΔT/Δx C)  Might be more complicated, nonlinear? D)  I don’t think it should depend on Δx.

dx

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x

dx

A

TH TC


What does H(x,t) depend on? We have concluded (so far)

H (x, t)∝ ∂ T (x, t)∂ x

Are we done?

2- 55

How does the prop constant depend on the area , A? A)  linearly B)  ~ some other positive power of A C)  inversely D)  ~ some negative power of A E)  It should be independent of area!

�

H(x, t)∝ ∂ T(x, t)∂x

Heat flow (H = Joules passing by/sec):

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Thermal heat flow H(x,t) has units (J passing)/sec

If you have H(x,t) entering on the left, and H(x+dx,t) exiting on the right, what is the energy building up inside, in time dt?

x

dx A

A)  H(x,dt)-H(x+dx,dt) B)  H(x+dx,t+dt)-H(x,t) C)  (H(x,t)-H(x+dx,t))dt D)  (H(x+dx,t)-H(x,t))/dt E)  Something else?! (Signs, units, factor of A, ...?)

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In steady state, in 1-D: solve for T(x)

∇2T = 1α 2

∂T∂t

x1 x2

T1 T2

2- 58

When solving T(x,y)=0, separation of variables says: try T(x,y) = X(x) Y(y) i)  Just for practice, invent some function T(x,y) that is manifestly of this form. (Don’t worry about whether it satisfies Laplace's equation, just make up some function!) What is your X(x) here? What is Y(y)? ii) Just to compare, invent some function T(x,y) that is definitely NOT of this form. Challenge questions: 1) Did your answer in i) satisfy Laplace’s eqn? 2)  Could our method (separation of variables) ever

FIND your function in part ii above?

�

∇2

2- 59

When solving T(x,y)=0, separation of variables says try T(x,y) = X(x) Y(y). We arrived at the equation f(x) + g(y) = 0 for some complicated f(x) and g(y)

�

∇2

Invent some function f(x) and some other function g(y) that satisfies this equation. Challenge question: In 3-D, the method of separation of variables would have gotten you to f(x)+g(y)+h(z)=0. Generalize your “invented solution” to this case.

2- 60

When solving T(x,y)=0, separation of variables says try T(x,y) = X(x) Y(y). We arrived at

�

∇2

�

d2X(x)dx 2

= cX(x)

andd2Y (y)dy 2

= −cY (y)

Write down the general solution to both of these ODEs!

Challenge: Is there any deep ambiguity about your solution?

2- 61

In part B of the Tutorial, you have f(x) = Csin(kx) + D cos(kx), with boundary conditions f(0)=f(L)=0. Is the f(x) you found at the end unique? A)  Yes, we found it. B)  Sort of – we found the solution, but it involves one

completely undetermined parameter C)  No, there are two very different solutions, and we

couldn’t choose! D)  No, there are infinitely many solutions, and we

couldn’t choose! E)  No, there are infinitely many solutions, each of

which has a completely undetermined parameter!

2- 62

__________________________________________ Question for you: Given the ODE, Which of these does the sign of “c” tell you? A)  Whether the solution is sines rather than cosines. B)  Whether the sol’n is sinusoidal vs exponential. C)  It specifies a boundary condition D)  None of these/something else!

d 2X(x)dx2

= cX(x)

2- 63

Last class we got to a situation where we had two totally unknown/unspecified functions a(x) and b(y), All we knew was that (for all x and all y) a(x) + b(y) = 0

What can you conclude about these functions?

A)  Really not much to conclude (except b(y)=-a(x) ! ) B)  Impossible, it’s never possible to solve this equation! C)  The only possible solution is the trivial one,

a(x)=b(y)=0 D)  a(x) must be a constant, and b(y)= -that constant. E)  I conclude something else, not listed!

2- 64

When solving T(x,y)=0, separation of variables says try T(x,y) = X(x) Y(y). We arrived at

�

∇2

�

d2X(x)dx 2

= cX(x)

andd2Y (y)dy 2

= −cY (y)

Write down the general solution to both of these ODEs!

Challenge: Is there any ambiguity about your solution?

2- 65

x=L

y=H

T=0

y=0 x=0

T=0

T=0

T=t(x)

Rectangular plate, with temperature fixed at edges:

Written mathematically, the left edge tells us T(0,y)=0. Write down analogous formulas for the other 3 edges. These are the boundary conditions for our problem

∇2T (x, y) = 0

2- 66

In part B of the Tutorial, you are looking for X(x) (we’re calling if f(x) here), f(x) = Csin(kx) + D cos(kx), with boundary conditions f(0)=f(L)=0. Is the f(x) you found at the end unique? A)  Yes, we found it. B)  Sort of – we found the solution, but it involves one

completely undetermined parameter C)  No, there are two very different solutions, and we

couldn’t choose! D)  No, there are infinitely many solutions, and we

couldn’t choose! E)  No, there are infinitely many solutions, each of

which has a completely undetermined parameter!

2- 67

x=L

T=0

y=0 x=0

T=0

Tè0

T=f(x)

Semi-infinite plate, with temp fixed at edges:

When using separation of variables, so T(x,y)=X(x)Y(y), which variable (x or y) has the sinusoidal solution? A)  X(x) B) Y(y) C) Either, it doesn’t matter D) NEITHER, the method won’t work here E) ???

∇2T (x, y) = 0

2- 68

We are solving T(x,y)=0, with boundary conditions: T(x,y)=0 for the left and and right side, and “top” (at ∞) T(0,y)=0, T(L,y)=0, T(x,∞)=0. The fourth boundary is T(x,0) = f(x) What can we conclude about our solution X(x)? �

∇2

A)  Cannot contain sin(ky) term B)  Cannot contain cos(ky) term C)  May contain both sin(ky) and cos(ky) terms D)  Must contain both sin(ky) and cos(ky) terms E)  Must contain both e-ky and e+ky terms

2- 69

We are solving T(x,y)=0, with boundary conditions: T(x,y)=0 for the left and and right side, and “top” (at ∞) T(0,y)=0, T(L,y)=0, T(x,∞)=0. The fourth boundary is T(x,0) = f(x) What can we conclude about our solution Y(y)? �

∇2

A)  Cannot contain e-ky term B)  Cannot contain e+ky term C)  Cannot contain either e-ky or e+ky terms D)  Must contain both e-ky and e+ky terms E)  ???

2- 70

Using 3 out of 4 boundaries, we have found Tn(x,y) = Ansin(n π x/L) e-nπ y/L

Question: Is ALSO a solution of Laplace’s equation? A) Yes B) No C) ???? �

T(x,y) = An sin(nπx /L) e−nπy /L

n=1

∞

∑

2- 71

Using 3 out of 4 boundaries, we have found Tn(x,y) = Ansin(n π x/L) e-nπ y/L Using the bottom boundary, T(x,0)=f(x), we can compute all the An’s: In terms of these (known) constants, what then is the complete final answer for T(x,y)?

�

An =2L

f (x)sin(nπx /L)dx0

L

∫

2- 72

Using 3 out of 4 boundaries, we have found Using the bottom (4th) boundary, T(x,0)=f(x), Mr. Fourier tells us how to compute all the An’s: �

An =2L


L

∫�


n=1

∞

∑

And we’re done!

2- 73

Using all 4 boundaries, we have found where Now suppose f(x) on the bottom boundary is T(x,0)=f(x) = 3sin(5 π x/L) What is the complete final answer for T(x,y)?

�


n=1

∞

∑

�

An =2L


L

∫

0 0

Tè0

3sin(5π x/L)

2- 74

x=L

y=H

T=0

y=0 x=0 T=0

T=0

T=t(y)

Rectangular plate, with temperature fixed at edges:

When using separation of variables, so T(x,y)=X(x)Y(y), which variable (x or y) has the sinusoidal solution? A)  X(x) B) Y(y) C) Either, it doesn’t matter D) NEITHER, the method won’t work here E) ???

�

∇2T(x,y) = 0

2- 75

x=L

y=H

T=0

y=0 x=0 T=0

T=t(y)

A)  i) k=n π/H, ii) A=-B B)  i) k=n π/L, ii) D=0 C)  i) A=-B, ii) k=n π/H D)  i) D=0, ii) k=n π/L E) Something else!!

Trial solution: T(x,y)=(Aekx+Be-kx)(Ccos(ky)+Dsin(ky))

Applying the boundary condition T=0 at i) y=0 and ii) y=H gives (in order!)

T=0

�

∇2T(x,y) = 0

2- 76

x=L

y=H

T=0

y=0 x=0 T=0

T=t(y)

A)  i) k=n π/H, ii) A=-B B)  i) k=n π/L, ii) D=0 C)  i) A=-B, ii) k=n π/H D)  i) D=0, ii) k=n π/L E) Something else!!



T=0

�

∇2T(x,y) = 0

i) C=0 ii) k = nπ/H

2- 77

x=L

y=H

T=0

y=0 x=0 T=0

T=t(y)

A)  i) A=-B, ii) k=n π/H B)  i) D=0, ii) k=n π/H C)  i) C=0, ii) k=n π/H D)  i) C=0, ii) k=n π/L E) Something else!!



T=0

�

∇2T(x,y) = 0

2- 78

A)  An=0 B)  Bn=0 C)  An=Bn D)  An=-Bn E)  Something entirely different!

Trial solution: Tn(x,y)=(Anenπx/H+Bne-nπx/H)(sin nπy/H)

Applying the boundary condition T(0,y)=0 gives...

x=L

y=H

T=0

T=0

T=t(y)

T=0

�

∇2T(x,y) = 0

2- 79

A)  Determines (one) An B)  Shows us the method of separation of v’bles failed in

this instance C)  Requires us to sum over n before looking for An’s D)  Something entirely different/not sure/...

Trial solution: Tn (x,y)=Ansinh(nπx/H)sin(nπy/H)

Applying the boundary condition T(L,y)=t(y) does what for us...

Recalling sinh(x)=½(ex-e-x)

x=L

y=H

T=0

T=0

T=t(y)

T=0

�

∇2T(x,y) = 0

2- 80

Trial solution:

What is the correct formula to find the An’s?

�

A) An =2L

t(y)sin(nπy/L)dy0

L

∫

B) An =2L

t(y)sin(nπy/H)dy0

L

∫

C) An =2H

t(y)sin(nπy/L)dy0

H

∫

D) An =2H

t(y)sin(nπy/H)dy0

H

∫E) Something entirely different!

x=L

y=H

T=0

T=0

T=t(y)

T=0

�

∇2T(x,y) = 0

�

T(x,y) = An sinh(nπx/H) sin(nπy/H)n=1

∞

∑

2- 81

Trial solution:

x=L

y=H

T=0

T=0

T=t(y)

T=0

�

∇2T(x,y) = 0

�

t(y) = T(L,y) = An sinh(nπ L/H) sin(nπy/H)n=1

∞

∑�


∞

∑

Right b’dry:

2- 82

�

bn =2H

t(y)sin(nπy/H)dy0

H

∫

x=L

y=H

T=0

T=0

T=t(y)

T=0

�

∇2T(x,y) = 0

�

T(L,y) = An sinh(nπ L/H)bn

sin(nπy/H)n=1

∞

∑

�

t(y) = bn sin(nπy/H)n=1

∞

∑

�

An sinh(nπL/H) =2H

t(y)sin(nπy/H)dy0

H

∫

Right b’dry:

Which means

2- 83 x=L

y=H

T=0

T=0

T=t(y)

T=0

�

∇2T(x,y) = 0

�

An =1

sinh(nπL/H)2H

t(y)sin(nπy/H)dy0

H

∫

Solution (!!) :

�


∞

∑

with:

2- 84 x=L

y=H

T=0

T=0

T=100

T=0

�

∇2T(x,y) = 0

�

An =1

sinh(nπL/H)2H

t(y)sin(nπy/H)dy0

H

∫

Solution (!!) :

�


∞

∑

with:

If e.g. t(y)=100° (a constant)...

2- 85

T=0

T=0 T=0

T=100

2- 86

x=L

y=H

T=0

T=0

T=f(y)

T=0

�

∇2T1(x,y) = 0

x=L

y=H

T=0

T=0

T=g(x)

T=0

�

∇2T2(x,y) = 0

�

with An =1

sinh(nπL/H)2H

f(y)sin(nπy/H)dy0

H

∫

�

T1(x,y) = An sinh(nπx/H) sin(nπy/H)n=1

∞

∑

How would you find T2(x,y)?

2- 87

x=L

y=H

T=0

T=0

T=f(y)

T=0

�

∇2T1(x,y) = 0

x=L

y=H

T=0

T=0

T=g(x)

T=0

�

∇2T2(x,y) = 0

�

with An =1

sinh(nπL/H)2H

t(y)sin(nπy/H)dy0

H

∫

�


∞

∑

�

with A'n =1

sinh(nπH /L)2L

g(x)sin(nπx/L)dx0

L

∫�

T2(x,y) = A'n sinh(nπy/L) sin(nπx/L)n=1

∞

∑

2- 88

x=L

y=H

T=0

T=0

T=f(y)

T=0

�

∇2T1(x,y) = 0

x=L

y=H

T=h(y)

T=0

T=0

T=0

�

∇2T3(x,y) = 0

�

with An =1

sinh(nπL/H)2H

f(y)sin(nπy/H)dy0

H

∫

�


∞

∑


2- 89

x=L

y=H

T=0

T=0

T=f(y)

T=0

�

∇2T1(x,y) = 0

x=L

y=H

T=h(y)

T=0

T=0

T=0

�

∇2T3(x,y) = 0

�

with An =1

sinh(nπL/H)2H

f(y)sin(nπy/H)dy0

H

∫

�


∞

∑

Just swap x with (L-x) (!)

2- 90

x=L

y=H

T=0

T=0

T=f(y)

T=0

�

∇2T1(x,y) = 0

x=L

y=H

T=0

T=0

T=g(x)

T=0

�

∇2T2(x,y) = 0

x=L

y=H

T=0

T=0

T=f(y)

�

∇2T4 (x,y) = 0

T=g(x)


2- 91

x=L

y=H

T=0

T=0

T=f(y)

T=0

�

∇2T1(x,y) = 0

x=L

y=H

T=0

T=0

T=g(x)

T=0

�

∇2T2(x,y) = 0

x=L

y=H

T=0

T=0

T=f(y)

�

∇2T4 (x,y) = 0

T=g(x)

T4(x,y) = T1(x,y) + T2(x,y)

Would this work?

A)  sweet! B)  No, it messes up Laplace’s eqn C)  No, it messes up Bound conditions D)  Other/??

2- 92

x=L

y=H

T=j(y)

T=h(x)

T=f(y)

�

∇2T(x,y) = 0

T=g(x)

We have solved this!

2- 93

Fourier Transforms

2- 94

If f(t) is periodic (period T), then we can write it as a Fourier series: What is the formula for cn?

�

f (t) = cnen=-∞

∞

∑i nω t

�

A) f (t)e− i nω t

0

T

∫ dω B) f (t)e− i nω t

0

T

∫ dt

C) 1ω

f (t)e−i nω t

0

T

∫ dω D) 1T

f (t)e− i nω t

0

T

∫ dt

E) Something else/not sure?

2- 95

Fourier Transforms Fourier Series

�

f (t) = g(ω)e iω t

-∞

∞

∫

�

f (t) = cnen=-∞

∞

∑i nω 0 t

2- 96


�

f (t) = g(ω)e iω t-∞

∞

∫ d(what?)

�

f (t) = cnen=-∞

∞

∑i nω 0 t

A)  dx B)  dt C)  dω D)  Nothing is needed, just E)  Something else/not sure

�

f (t) = g(ω)eiω t0

∞

∫

2- 97

(limit as T gets long) Fourier Transforms

(period T = 2π/ω0) Fourier Series

f (t)∝ g(ω )eiω t-∞

∞

∫ d(what?)

�

f (t) = cnen=-∞

∞

∑i nω 0 t

A)  dx B)  dt C)  dω D)  Nothing is needed, just E)  Something else/not sure

f (t)∝ g(ω )eiω t0

∞

∫

2- 98


�

f (t) = g(ω)eiω t

-∞

∞

∫ dω

�

f (t) = cnen=-∞

∞

∑i nω 0 t

�

cn =1T

f (t)e−i nω 0 t

0

T

∫ dt

�

g(ω)∝ f (t)ei what?dwhat?-∞

∞

∫

�

A) f (t)e−i nω tdt-∞

∞

∫ B) f (t)e−iω tdt-∞

∞

∫C) f (t)e− i nω tdω

-∞

∞

∫ D) f (t)e−iω tdω-∞

∞

∫E) Something else/not sure?

2- 99


�


-∞

∞

∫ dω

�

f (t) = cnen=−∞

∞

∑i nω 0 t

�

cn =1T

f (t)e−i nω 0 t

0

T

∫ dt

�

g(ω) =12π

f (t)e−iω tdt-∞

∞

∫

2- 100

g(ω) is the Fourier Transform of f(t)

�


-∞

∞

∫ dω

�

g(ω) =12π

f (t)e−iω tdt-∞

∞

∫f(t) is the inverse Fourier Transform of g(ω)

2-

Consider the function

It is … A) zero B) non-zero and pure real C) non-zero and pure imaginary D) non-zero and complex

�

f (t) = e− t2 / b

50

�

f (t)e−iω tdt−∞

∞

∫

101

f(t)

t

What can you say about the integral

2-

We are planning to hold a help session the Monday of final exam week (5/7/12). This is the day before our final exam. We’d like to know when it would make sense to hold it. Please select the time that works best for you.

A)  10am-Noon B)  2pm-4pm C)  4pm-6pm D) I probably wouldn’t attend anyway E) Some other time works best for me.

2-

If f(t) is given in the picture, it's easy enough to evaluate

t +T0 -T0

1/(2T0) f(t)

�

g(ω) =12π


∞

∫Give it a shot! After you find a formula, is it... A)  real and even B)  real and odd C)  complex D)  Not sure how to do this... 103

2-

If f(t) is given in the picture, �

g(ω) =12π


∞

∫

104

A)  0 B) infinite C) 1/2π D) 1/(2πωT0) E) something else/not defined/not sure... �

lim(ω→0)

g(ω)?What is

�

=12πsinωT0ωT0

2-

If f(t) is given in the picture, �

g(ω) =12π


∞

∫

105

What is

A)  0 B) infinite C) 1/2π D) 1/(2πωT0) E) something else/not defined/not sure... �

lim(ω→∞ )

g(ω)?What is

�

=12πsinωT0ωT0

2-

If f(t) is given in the picture,

Describe and sketch g(ω)

�

g(ω) =12π


∞

∫

106

Challenge: What changes if T0 is very SMALL? How about if T0 is very LARGE?

�

=12πsinωT0ωT0

2-

Consider the function f(x) which is a sin wave of length L.

• Which statement is closest to the truth? A)  f(x) has a single well-defined wavelength B)  f(x) is made up of a range of wavelengths

42

�

f (x) = sin(kx), − L2

< x < + L2

0, elsewhere

⎧ ⎨ ⎪

⎩ ⎪

2- 108

What is the Fourier transform of a Dirac delta function, f(t)=δ(t)?

A)  0 B)  ∞ C)  1 D)  1/2π E)  e-iω

�

g(ω) =1

2πf (t)e−iω tdt

−∞

∞

∫ = ?

2- 109

What is the Fourier transform of a Dirac delta function, f(t)=δ(t-t0)?

�

g(ω) =1

2πf (t)e−iω tdt

−∞

∞

∫ = ?

E) Something else...

�

A) 12π

B) 12π

δ(ω)

C) 12π

e-iω t D) 12π

e-iω t 0

2- 110

The Fourier transform of

Sketch this function

�

g(ω) = σ 2πe−ω2σ 2 / 2

�

f (t) = e− t2 /(2σ 2 )

is

2- 111

What is the standard deviation of which is the Fourier transform of

A)  1 B)  σ C)  σ2 D)  1/σ E)  1/σ2

�

g(ω) = σ 2πe−ω2σ 2 / 2

�

f (t) = e− t2 /(2σ 2 )

2-

Compared to the original function f(t), the Fourier transform function g(ω)

A) Contains additional information B) Contains the same amount of information C) Contains less information D) It depends

112

2- 113

Match the function (on the left) to its Fourier transform (on the right)

2- 114

Solving Laplace’s Equation: If separation of variables doesn’t work, could use “Relaxation method”

∇2T (x, y) = 0

2- 115

Solving Laplace’s Equation:

∇2T (x, y) = 0

A handy theorem about any solution of this eq’n: The average value of T (averaged over any sphere) Equals the value of T at the center of that sphere.

2- 116


∇2T (x, y) = 0

1. Break region up in to a grid of points.

T=y

T=x(0,0) (5,0)

(0,5)

T=5

T=5

2- 117


∇2T (x, y) = 0


T=y

T=x(0,0) (5,0)

(0,5)

T=5

2- 118


∇2T (x, y) = 0


T=y

T=x(0,0) (5,0)

(0,5)

T=5

2- 119

2- 120

2- 121

phys2210 sp12 07 FourierandPDEs · Dirac delta function, PDEs, and Fourier transforms If we have...

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Transcript of phys2210 sp12 07 FourierandPDEs · Dirac delta function, PDEs, and Fourier transforms If we have...